Question

Prove that cos⁻¹$\frac{4}{5}$ + sin⁻¹$\frac{3}{\sqrt{34}}$ = tan⁻¹$\frac{27}{11}$.

Answer 1

Answer:

Step-by-step explanation:

$cos^{-1}\frac{4}{5} + sin^{-1}\frac{3}{\sqrt{34}} = tan^{-1}\frac{27}{11}$

as we know that

$cos^{-1}x=tan^{-1}(\frac{\sqrt{1-x^{2} } }{x} )\\\\cos^{-1}\frac{4}{5} =tan^{-1}(\frac{\sqrt{1-(\frac{4}{5})^{2} } }{\frac{4}{5}} )\\\\\\=tan^{-1}(\frac{\sqrt{1-(\frac{16}{25}) } }{\frac{4}{5}} )\\\\\\=tan^{-1}(\frac{\sqrt{(\frac{25-16}{25}) } }{\frac{4}{5}} )\\\\=tan^{-1}\frac{3}{4}$

by the same way sin inverse can be written in terms of tan

$sin^{-1}x=tan^{-1}(\frac{x}{\sqrt{1-x^{2} } } )\\\\sin^{-1}\frac{3}{\sqrt{34}} =tan^{-1}(\frac{\frac{3}{\sqrt{34}}}{\sqrt{1-(\frac{3}{\sqrt{34}})^{2} } } )\\\\\\=tan^{-1}(\frac{\frac{3}{\sqrt{34}}}{\sqrt{1-(\frac{9}{34}) } } )=tan^{-1}(\frac{\frac{3}{\sqrt{34}}}{\sqrt{(\frac{34-9}{34}) } } )=tan^{-1}\frac{3}{5}$

as we know that

$tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy} )$

$tan^{-1}\frac{3}{4} +tan^{-1}\frac{3}{5}=tan^{-1}(\frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4}\frac{3}{5}} )\\\\\\=tan^{-1}\frac{\frac{15+12}{20} }{\frac{20-9}{20} } \\\\\\=tan^{-1}\frac{27}{11}$

hence proved